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lukranit [14]
3 years ago
12

It takes 26.8 mL of a 0.0700 M NaOH standard solution to neutralize a 250 mL sample of lactic acid (C3H6O3). What mass of lactic

acid was dissolved in the sample?
Chemistry
1 answer:
Butoxors [25]3 years ago
4 0
There is 1 OH- in 1 molecule of NaOH. 
Also, there is 1 H+ in 1 molecule of lactic acid.

So the reaction is simple.
so just equate the moles
moles of OH- in NaOH = moles of H+ in lactic acid
26.8 x 0.07 = 250 x Mole of lactic
Moles of lactic = 0.0075

so mass = 0.0075 x 90.8 =  0.681 g

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What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter
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Answer : The final temperature of the solution in the calorimeter is, 31.0^oC

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First we have to calculate the heat produced.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of NaOH = 1.52 g

Molar mass of NaOH = 40 g/mol

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole

Now put all the given values in the above formula, we get:

44.5kJ/mol=\frac{q}{0.038mol}

q=1.691kJ

Now we have to calculate the final temperature of solution in the calorimeter.

q=m\times c\times (T_2-T_1)

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q = heat produced = 1.691 kJ = 1691 J

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Now put all the given values in the above formula, we get:

1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)

T_2=31.0^oC

Thus, the final temperature of the solution in the calorimeter is, 31.0^oC

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3 years ago
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